Let $x^3+y^2=24$. What is the value of $\dfrac{d^2y}{dx^2}$ at the point $(2,4)$ ? Give an exact number.
Notice that the equation defines $y$ implicitly—we don't have an explicit expression for $y$ in terms of $x$. So we will have to use implicit differentiation. If we differentiate the equation once, we will be able to get an expression for $\dfrac{dy}{dx}$. Then we can differentiate the equation again to get an expression for $\dfrac{d^2y}{dx^2}$. Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=-\dfrac{3x^2}{2y}$ Now we can differentiate $\dfrac{dy}{dx}$ to find $\dfrac{d^2y}{dx^2}$. $\dfrac{d^2y}{dx^2}=-\dfrac{12xy^2+9x^4}{4y^3}$ Finally, let's plug ${x=2}$ and ${y=4}$ into the expression we got: $\begin{aligned} \left.-\dfrac{12 x y^2+9 x^4}{4 y^3}\right\rvert_{({2},{4})}&=-\dfrac{12({2})({4})^2+9({2})^4}{4({4})^3} \\\\ &=-\dfrac{33}{16} \end{aligned}$ In conclusion, the value of $\dfrac{d^2y}{dx^2}$ at the point $(2,4)$ is $-\dfrac{33}{16}$.